Determine if $\sum_{n=0}^∞ \frac{n}{(2i)^n}$ is convergent, divergent

Question

$\sum_{n=0}^∞ \frac{n}{(2i)^n}$

I have been trying to work out how to do this as this involves $i$, usually we would use ratio test but I wasn't sure how to.

How would you work this out with derivative?

Answer 1

A series converges to a complex number if and only if its absolute value converges to a real number. Thus, let's look at the absolute value.

$$\Big|\sum_{n=0}^{\infty} \dfrac{n}{(2i)^n}\Big| \leq \sum_{n=0}^{\infty} \Big|\dfrac{n}{(2i)^n}\Big| = \sum_{n=0}^{\infty} \dfrac{n}{2^n}, $$

which converges, ensuring that the absolute value (and thus the series itself) of our original series converges by the comparison test.

Answer 2

Guide:

\begin{align} \sum_{n=0}^\infty \frac{n}{(2i)^n} &= \frac1{2i} \sum_{n=0}^\infty n \left( \frac1{2i}\right)^{n-1} \\ &= \frac1{2i} \sum_{n=0}^\infty\frac{d}{dx} x^n \mid_{x=\frac1{2i}}\\ &= \frac1{2i} \frac{d}{dx}\sum_{n=0}^\infty x^n \mid_{x=\frac1{2i}} \\ &= \frac1{2i}\frac{d}{dx} \left( \frac{1}{1-x}\right)|_{x= \frac1{2i}} \end{align}

Can you use geometric series to evaluate the quantity of interest?

Answer 3

Consider the following power series.

$$\sum_{n=0}^{\infty}a_nz^n$$

where $a_n=\frac{n}{2^n}$.

Radius of convergence of the above series is $\limsup_{n\to\infty}|\frac{a_{n}}{a_{n+1}}| $ and that turns out to be $2$

So the above series converges for $|z|<2$

Take $z=1/i$. What is $|1/i|?$ It is equal to $1$ and as we have shown above that the series converges for all $|z|<2$, in particular it converges for $z=1/i$.

Answer 4

Show that the series is absolutely convergent.

$\left |\dfrac{n}{(2i)^n} \right |= \dfrac{n}{2^n}.$

Root test.

$\lim_{n \rightarrow \infty} \sup \sqrt [n]{|a_n|}=$

$\lim_{n \rightarrow \infty} \sup_{k \ge n} (\sqrt[k]{\dfrac{k}{2^k}})= 1/2 <1$,

hence convergent.

Absolute convergence of the series implies convergence .